a^2n +b^2n=c^2n

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Rafe
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a^2n +b^2n=c^2n

Unread post by Rafe » Tue Nov 06, 2012 6:11 pm

Prove that if $a^{2n}+b^{2n}=c^{2n}$ where $a,b,c$ are positive,then $a^n+b^n>c^n$.Can you do it by geometry?
Last edited by *Mahi* on Tue Nov 06, 2012 10:34 pm, edited 2 times in total.
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Phlembac Adib Hasan
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Re: a^2n +b^2n=c^2n

Unread post by Phlembac Adib Hasan » Tue Nov 06, 2012 8:58 pm

Rafe wrote:PROVE THAT IF $a^{2n}+b^{2n}=c^{2n}$ WHERE $a,b,c$ ARE POSITIVE,THEN $a^n+b^n>c^n$. CAN YOU DO IT BY GEOMETRY
Use $L^AT_EX$ please, it makes our work a lot easier.
Let $a^n=x,b^n=y,c^n=z$. So the equation implies $x^2+y^2=z^2$
So from Pythagoras's theorem, it is possible to construct a right angled triangle with side lengths $x,y,z$ when $z$ is the hypotenuse. Now from triangle inequality, it easily follows $x+y>z$.
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